If f (x) = ^ - 1 ( 2 × 3 ^ x 1 + 9 ^ x ), then f ^ ' (- (1)/(2)) is equal to

Correct answer: A. 3 _ e 3 Given f(x) = si n ^ - 1 ( 2 × 3 ^ x 1 + 9 ^ x ) Let 3 ^ x = (t) ⇒ t = ^ - 1 ( 3 ^ x ) So, f ( x ) = si n ^ - 1 ( 2 (t) 1 + ta n ^ 2 (t) ) We know that, ( 2t ) = 2 ( t ) 1 + ^ 2 ( t ) ⇒ f ( x ) = si n ^ - 1 ( (2t)) ∴ f ( x ) = 2t = 2ta n ^ - 1 ( 3 ^ x ) ⇒ (df(x))/(dx) = 2 1 + ( 3 ^ x ) ^ 2 × 3 ^ x . lo g _ e 3 At x = (1)/(2), (df)/(dx) = 2 1 + (3(1)/(2)) ^ 2 × 3(1)/(2) . lo g _ e 3 = (1)/(2) × 3 × _ e 3 = 3 × _ e 3

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

If f (x) = ^ - 1 ( 2 × 3 ^ x 1 + 9 ^ x ), then f ^ ' (- (1)/(2)) is...

f

\left(x\right) = \sin^{- 1}

\left(\dfrac{2 \times {3}^{x}}{1 + {9}^{x}}\right),

then

{f}^{'}\left(- \dfrac{1}{2}\right)

is equal to

Options

A $\sqrt{3}\log_{e}\sqrt{3}$
B $- \sqrt{3}\log_{e}\sqrt{3}$
C $- \sqrt{3}\log_{e}3$
D $\sqrt{3}\log_{e}3$

Correct answer

A. $\sqrt{3}\log_{e}\sqrt{3}$

Step-by-step solution

Given

f\left(x\right) = si{n}^{- 1}\left(\dfrac{2 \times {3}^{x}}{1 + {9}^{x}}\right)

Let

{3}^{x} = \tan \left(t\right) ⇒ t = \tan^{- 1}\left({3}^{x}\right)

So,

f ( x ) = si{n}^{- 1}\left(\dfrac{2\tan \left(t\right)}{1 + ta{n}^{2}\left(t\right)}\right)

We know that,

\sin ( 2t ) = \dfrac{2\tan ( t )}{1 + \tan^{2} ( t )}

⇒ f ( x ) = si{n}^{- 1}\left(\sin \left(2t\right)\right)

∴ f ( x ) = 2t = 2ta{n}^{- 1}\left({3}^{x}\right)

⇒ \dfrac{df\left(x\right)}{dx} = \dfrac{2}{1 + {( {3}^{x} )}^{2}} \times {3}^{x} . lo{g}_{e}3

x = \dfrac{1}{2}, \dfrac{df}{dx} = \dfrac{2}{1 + {\left(3\dfrac{1}{2}\right)}^{2}} \times 3\dfrac{1}{2} . lo{g}_{e}3

= \dfrac{1}{2} \times \sqrt{3} \times \log_{e}3

= \sqrt{3} \times \log_{e}\sqrt{3}

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